Example- Motional emf
Motional electrode motor force. Let’s consider a magnetic field, a uniform magnetic field pointing into the plane. B is uniform and pointing into the plane and here inside of this magnetic field region we have a rectangular conducting loop which is being pulled out of this magnetic field region at a constant velocity. Therefore it is moving out at a constant velocity. Let’s try to analyze this system. Well, as we pull out this conducting loop, rectangular loop, out side of this magnetic field region the number of field lines passing through the area surrounded by this loop will get lesser and lesser. It means that the flux through the area, surrounded by this loop will decrease. Since the magnetic flux is changing.
Then we’re going to end up with a induced electron motor force from Faraday’s Law. And that electron motor force will cause an induced current and that current will appear along this loop from Lenz’s Law . Such as it will oppose it’s course. It’s course is the decrease in flux which caused as a result of movement of this loop outside of this field region. So the induced EMF is going to show up in such a way that it will oppose it’s course, the course decreases in flux therefore it will try to compensate that decrease and the only way that it can do that, by generating a magnetic field which will be in the same direction with the direction of this external magnetic field. So the induced current’s magnetic field is going to be pointing into the plane through the area surrounded by this coil. Using Right-Hand rule to be able to have magnetic field lines, which are going into the plane through the area surrounded by this coil; holding the right hand finger’s direction in that rotational direction. That is a clockwise direction. The thumb will give us the direction of the associated current and in this case that current therefore has to be flowing in clockwise direction along the slope.
Now as we move this loop, at constant velocity outside this magnetic field region. Therefore we will end up with this induced current, but once the current appears then we will end up with another case and that is we will have external magnetic field D and current carrying Y inside of this region. In other words this magnetic field will exert force on every segment of this loop inside of this magnetic field region and that force as you recall is equal to integral of IDL cross B if we look at the direction of that force for these segments for the left hand side segment IDL is going to be pointing in outward direction. In the direction of flow of current. So if we choose an incremental element like this it will be pointing in outward direction and magnetic field is into the plane. IDL cross B will give us an incremental force along the segment pointing to the left and along the, all IDLs through out the length of this segment they will be all pointing outside. And if you add them vectorially we are going to end up with a net force of let’s say F sub B pointing to the left on the segment. In a similar way if we look at the net force on the upper segment of this loop IDL is going to be pointing to the right B is into the plane IDL cross B will generate a force in this case in outward direction. so lets mark these forces. Like this therefore this one, let’s call this one Fb prime for the upper segment pointing up like this and for the lower segment IDL is going to be pointing to the left and IDL cross B; again by applying Right-Hand Rule, will generate a force in downward direction. Since the magnitude of the current is the same for upper and as well as the lower segment and they’re equal in length wise. F sub D primes will be equal to each other magnitude wise. And similarly this ant is also in the magnetic field region which will be under the influence of this force of F sub B pointing to the left. When look at the right hand side region though. There is no or for this segment there is no magnetic field in this region. I mean external magnetic field.
Of course we have the magnetic field of the induced current, but as we know the magnetic field of the current it’s self will not exert or generate any force on that current. So this ant is not going to be under the influence of any force exerted by this external magnetic field over here. So under this condition then all though these two forces F b primes acting along the same line with opposite directions and equal magnitude will cancel one another. But there is no force to balance out F sub B which is pointing to the left. So in order to keep the, this loop, conducting loop to move at this constant velocity the net force acting on the object should be equal to zero from Newton’s first law. Therefore we have to balance this magnetic force with a force in the opposite direction, equal magnitude. So it means that we have to apply a force at this end an external force such that it has to be equal in magnitude wise to this magnetic force. So the loop can continue to move at a constant velocity V. So we can make a note over here F sub B primes will cancel and in order to keep the conducting loop to move at constant velocity and if external has to be applied. Such that, magnitude wise, if external should be equal to F sub B. Well if we look at the magnitude of this force F sub B magnitude is equal to integral of IDL. Let’s first write it down in vector form. The magnitude is going to be equal to integral of IDL of cross B magnitude. Which is going to give us F sub B is equal to the magnitude expression for this cross product is IDL magnitude times D magnitude, the magnitude of the magnetic field, times SINE of the angle between vectors. If we look at our diagram IDL is on the plane and in up ward direction where as D is into the plane. Therefore the angle between these two vectors is 90 degrees. SINE 90 is just one current and magnetic field. These are constant. We can take it outside the integral and integral is going to be taken along the length of this segment. We can give dimension to the segment, let’s say that, that distance is equal to L.
Therefore integral is going to be taken over the length. In other words the boundary will go from 0 to L. So the magnitude of the magnetic force will be equal to ID times integral of DL integrated from 0 to L and that is going to give us, just the length of that segment which is L. IDL will be the magnitude of the magnetic force. So we have to apply an external force; has to be equal to the magnitude of this force which will be IDL. Of course this force has to be applied in, on the other segment across from this one. When in opposite direction to this so that the the net force acting on this loop will be zero as a result of that from Newton’s first law. It will continue to move at constant velocity. Alright, now let’s try to determine the induced current. Well from Ohm’s law, current is equal to voltage divided by resistance. Therefore to be able to obtain the current we need to first determine the induced electrode motor force. Let’s say, let are be the resistance of this loop. In order to determine the induced voltage we will apply Faraday’s Law. Which was minus D Phi B over D t. Negative of the rate of change of magnetic field. Here again this negative sign is directly associated with the Lanz’s law and it simply states that the flow direction of induced current is such that it opposes it’s course. OK, Phi sub B is the magnetic flux and it is defined as the magnetic field vector, both with incremental area vector integrated over an open surface. In our case if we call the length of the loop which is inside of the external magnetic field region as X. Then at an instance of time the flux, magnetic flux is going to be equal to B magnitude, DA magnitude times COSINE of the angle within these two vectors. Here we now know that the surface is surrounded by a current loop.
Using Right-Hand Rule we hold our right hand fingers in the the direction of flow of current which is in clockwise direction and keeping the thumb in up position or open position then we will get the corresponding area vector. And for every incremental area along the surface then that direction will be pointing into the plane. So B, the angle between B and DA will be zero degrees. And COSINE of zero is just one. The magnetic field magnitude is constant we can take it outside of the integral and therefore we end up with the internal of DA, integrated overall the region of this loop inside of the magnetic field region and that is basically this blue region. And the area of that region once we add all those incremental area vectors to one another becomes equal to it’s width times length L times X. Therefore at an instant of time magnetic flux becomes equal to B times L time X. This magnetic field will change as this distance X changes. And incremental change in magnetic field will occur when ever we have an incremental change in DX occurs. If we look at the rate of change of this quantity D Phi B over DT that will be equal to D is constant, L is constant times DX over DT. But rate of change of this displacement will be equal to velocity and the magnitude of that will be equal to the speed of this loop as it moves out from this external magnetic field region. Therefore this quantity is going to be equal to just V. Now D Phi B over DT is by definition is induced electrode motor force then epsilon becomes equal to D times L time V. Again of course we have a negative sign over here, but that negative sign as I mentioned earlier is directly associated which direction that the current is going to be flowing through this loop. The magnitude of the induced electrode motor force will be equal to this quantity. As we can see over here it is equal to the product B of product of magnetic field, the width of the slope L and times the speed of this loop. And since it is related to the velocity of the moving loop this one is also referred as motional electrode motor force.
OK now we can calculate the current by using Ohm’s law R is equal to voltage divided by the resistance therefore epsilon divided by R is going to give us DLV over R. Then if we go back to our force expression over here it is going to be equal to, force was equal to I times B times L and by using the explicit form of the current we can also express this force, the magnitude of the force we have to apply to the loop in order to move at constant velocity DLV over R times B times L this is the current part and that becomes equal to B square, L square, V divided by R.
All right, as you recall work done was defined as incremental work done as defined as force vector dotted with displacement vector. And from work energy CRM this quantity is equal to change in kinetic energy or if the force field is a conservative force field in that case it was negative of the change in potential energy. Now if we look at the rate of work done. Which is energy per unit time. That will be equal to, if I write this down right inside exponent form since force and the displacement vector are in the same direction, product of their magnitudes times COSINE of zero will be the explicit form of this work expression and the rate of change of work will be equal to F times DX over DT. And again DX over DT is just the velocity so the product of force magnitude of force times the magnitude of velocity speed becomes equal to energy per unit time. And that is nothing but power. So here by applying this force of external, and causing the loop to displace by a distance of DX we’re supplying certain amounts of energy in every second and that is a power that we supply. So we can express power supplied to the system as F times V in explicit form we have just found that the force is equal to B square, L square V over R and times V is going to give us how much energy that we supplied to the system per unit time. Well if we are supply this much of energy to this rectangular room by moving it under the influence, I mean by causing it to move at a constant velocity let’s see where this energy is consumed in every second.
Let’s go back to our diagram, now apply this as if external we’re just balancing the force, magnetic force therefore the net force is becoming zero. Acting on this loop then it will, it is continuing to move at constant velocity of V. As we do this we’re doing some work and we’re supplying some energy to the system and the amount of energy we are supplying to the system per unit time in explicit form is equal to this expression: B squared L squared over R. And as we do this a current is appearing along the loop and that current is flowing through the resistance of this loop so it means that, that energy will be dissipated in the form of heat. And if you recall energy dissipated per unit time through a resistor is a result of current was equal to square of the current times that resistances. Well, power dissipated then will be equal to,we have found earlier that current flowing through the loop is BL , V or R. Square of that will be B square, L square or R square, therefore this is R square and if we needed to multiply that by the resistances to be able to get the power dissipated this R and that R squared in the denominator will cancel therefore power dissipated will be equal to B square, L square, V square over R. If you compare power supply, which is this quantity, and power dissipated we will see that these two quantities are equal to one another. And that’s what we expect directly from the conservation of energy principle